How do I go about solving 3^{x^2+x}=9. Thanks in advance
How do I go about solving 3^{x^2+x}=9. Thanks in advance
Hint: Write 9 as 3^2
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
Let's get a new equation, first. After subsituting 3^2 for 9, we have this:
3^(x^2 + x) = 3^2
This equation shows two powers of 3 set equal to one another. Can you use a basic property of exponents, to write another equation where x does not appear in any exponents?
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
That's part of the answer.
Did you write (and maybe solve) a new equation, as I suggested? If so, may I see the equation and your work?
If you did not solve an equation, how did you get x = 1?
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
Here's the basic property of exponents that I have in mind:
Given b^m = b^n, then m = n
In other words, if two powers of the same base are equal, then the exponents must be equal. Here are some examples:
14^2 = 14^z means z = 2
A^(4y) = A^(y - 5) means 4y = y - 5
(3/4)^(t^3/17) = (3/4)^(4t^5) means t^3/17 = 4t^5
If you apply this property to the given equation in your exercise (after substituting 3^2 for 9), you'll get a basic quadratic equation to solve for x. (There are two solutions.)
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
Please reply showing how you followed through on the steps and hints you were provided. You created the equation they'd almost given you, you solved this using the Quadratic Formula, you got the two values, and... how did you get that "x=1" is the only answer?
Please be complete. Thank you!
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